
Class / .g J 3 
Book._ 
Copyright ]^°_ 



COPYRIGHT DEPOSnV 



i'-'W 



Constructive Drawing. 

A TEXT-BOOK FOR HOME INSTRUCTION, HIGH SCHOOLS, MANUAL TRAINING HIGH 
SCHOOLS, TECHNICAL SCHOOLS AND UNIVERSITIES. 

ARRANGED AND PUBLISHED BY 

HERMAN HANSTEIN 

Instructor of Drawing A. G.Lane Manual Training High School, Director of Art and Technical Drawing: Departments. Chicago 

Mechanics Institute, Chicago Business College, formerly Supervisor of Art and Constructive 

Drawing Departments Chicago High Schools. 




KEUPPEL & 
ESSER CO. 



THIRD EDITION. 

GEOMETRICAL CONSTRUCTIONS. 
CHICAGO 1908 



NEW YORK 

CHICAGO 

ST. LOUIS 

SAN FRANCISCO 



AUG t7,^a08 

OUA$S (X^ _JJ^ XKu 

I »4- i^ s i, 

COPY B. 



^V 



W^ 



oa,.2^^^ 



ENTERED ACCORDING TO ACT OF CONGRESS, TN THE YEAR 1906 



3Y HERMAN HANSTEI 



IN THE OFFICE OF THE LIBRARIAN OF CONGRESS, AT WASHINGTON, 



PREFACE TO THIRD EDITION 



A T the request of former assistants and pupils I have compiled and 
■'- *■ revised these problems, as a help and for home instruction. This 
work represents the first year's course, that has been followed the past 
twenty-five years in the Chicago High Schools and in the Drawing Depart- 
ments of the Chicago Mechanics Institute. 

A practical experience of seventeen years in office and shop and his 
occupation as teacher during the past thirty years have given the author 
such experience and judgment as to select only such problems as are of 
practical importance to those who follow architectural, mechanical and 
engineering vocations, as well as problems which are indispensible to 
manufacturing and industrial pursuits. 

The author feels very grateful for the manner in which his former 
editions were received and hopes this revised third edition will meet with 
increased favor. 

HERMAN HANSTEIN, 
Cliicago, III., July 1908. 361 Mohawk Street 



INDEX. 



Tools, implements and their applications Figs. 1 to 7, Plate 1 

Alphabets Plate 2 

Block Alphabets Plate 3 

Bet squares and their applications Figs. 1 to 4, Plate 3 

Comers for borders Fig. 5, Plate S 

DEPISITIONS. 

Lines and angles Text opposite Plate 2 

Planes and surfaces " '" Plate 3 

Polygons " " Plate 3 

Circle " " Plate 3 

OONSTROOTION OP PERPBNDICnLARS. 

To construct perpendiculars Nos. 1 to 6, Figs. 1 to 6. Plate i 

To construct perpendiculars Nos. 7 and 3, Figs. 1 and 2, Plate 6 

DIVISIONS OF LINES. 

To divide lines Nos. 8 to 12, Figs. 3 to 6, Plate 6 

SOLUTIONS OF ANGLES. 

To construct and divide angles Nos. 13 to 18, Figs. I to 4, Plate 6 

SOLUTIONS OF TRIANGLES. 

To construct triangles Nos. 17 to 20, Figs. B and 6, Plate 6 and Figs. 1 and 2, Plate 7 

PROPORTIONAL LINES. 

To construct proportional lines Nos. 21 to 84, Figs. 3 to 6, Plate 7 

POLYGONS. 

To construct regular polygons on given bases Noe. 25 to 39. Figs. 1 to 4, Plate 8 

To construct equivalent polygons Nos. 33 to 40, Figs. 6 and 6, Plate Sand Figs. 1 to 6, Plate 9 

TRANSFERRING POLYGONS. 

To construct polygons equal to a given one Nos. 41 to 46, Figs. 1 to 6. Plate 10 and Figs. 1 to 6, Plate 11 

REDUCTION OK ENLARGEMENT OF POLYGONS. 

To construct polygons proportional in area or outline and 

similar to given ones Nos. 47 to 52, Figs. 1 to 11, Plate 12 and Figs. 1 to 5, Plate 13 

SCALES. 

To construct scales Nos. 53 to 56, Figs. 6 au J 6 A, Plate 13 and Figa. 1 to B, Plate 14 

CIKCLEa. 

To divide circles Nos. 57 to 60, Figs. 1 to 5, Plate 15 

RECTIFICATION OF ARCS. 

To rectify arcs Nos. 61 and 62, Fig. 6, Plate 15 and Fig. 1, Plate 16 



INDEX.— Continued 

TANGENTS. 

To construct tangents Nos. 63 to 66, Figs. 2 to 6, Plate 16 

TANGENTIAL CIRCLES. 

To construct tangential circles Nos. 67 to 81, Figs. 1 to 6, Plate 17, Fi^s. 1 to 6, Plate 18, Figs. 1 to 4, Plate 19 

GOTHIC AND PERSIAN ARCHES. 

To construct Gothic and Persian arches Nos. 82 to 87, Figi* 6 and 6, Plate 19, Figs. 1 to 4, Plate 20 

KGG-LINES. 

To construct egg-lines Nos. 88 and 89, Figs. 6 and 6, Plate 80 

OVALS. 

To construct ovals Nos. 90 to 93, Figs. 1 to 4, Plate 21 

ARCHES. 

To construct arches Nos. 94 to 98, Figs. B and 6, Plate 21, Figs. 1,2 and 4, Plate 22 

ASCENDING ARCHES. 

To construct ascending arches Nos. 99 to 102, Figs. 3 and 6, Plate 22, Figs. 2 and 3, Plate 23 

SPIRALS. 

To construct spirals Nos. 103 to 106, Fig. 1, Plate 23, Figs. 1 to 3, Plate 24 

CAM LINES OR ARCHIMEDEAN SPIRALS. 

To construct cams Nos. 107 to 110, Figs. 1 to 4, Plate 25 

CONIC SECTIONS— ELLIPSE. 

To construct ellipses Nos. Ill to 120, Figs. 6 and 6, Plate 25, Figs. 1 to 8, Plate 26 

PARABOLA, 

To construct parabolas Nos. 121 to 126, Figs. 1 to 6, Plate 27 

HYPERBOLA. 

To construct hyperbolas Nos. 127 to 129, Figs. 1 to3, Plate 28 

GEAR LINES— CYCLOID AND EVOLUTB. 

To construct an evolute No 130, Fig. 4, Plate 28 

To construct a cyloid Nos. 131 to 133, Figs, land 2, Plate 29 

To construct an epicycloid and hypocycloid (epitrochoid and 

hypotroohoid) Nos. 184 to 138, Figs. 1 to 3, Plate 30, Figs. 1 and 2, Plate 31 

APPLIED CONSTRUCTIONS IN ARCHITECTURE AND MECHANICS. 

To construct an ornamented Gothic arch No. 139, Fig. 1, Plate 32 

To construct a pair of "spur" whet'ls No. 140, Fig. 1, Plate 83 



Necessary Tools, Implements and Their Application. 



Fig. 1, Plate!.— .4. drawing-board, made of well- 
seasoned white pine, poplar (whitewood) or 
basswood, the lightest of our woods, answers 
this purpose best, as these woods are evenly 
grained and do not offer great obstruction to 
thumb-tacks, by which the drawing-paper is 
fastened to the board. 

The under surface of this board should be 
provided with two parallel dovetailed grooves, 
3 or 4 inches from edges O and O' and right- 
angled to the grain of the wood, to receive not 
too tightly fitting cleats, at which the board may 
shrink, to prevent its splitting. The cleats 
therefore should not be glued in the grooves to 
receive them. 

When one draws with the right hand, the 
straightedge, called T square (T), and triangle 
S, called set square, are operated with the left 
hand, and when one draws with the left hand 
the set and T square are operated with the right 
hand. 
The T is used only on one edge of the hoard. 

Pigs. 1 and 2, Plate 3.— Set squares (Triangles). — 
One set square of 30" and GO" and one of 45" 
(degrees) are required, as shown in Plate 3 and 
these should be tested for accuracy before ad- 
mitted to practical use. 



I'est. — Place the set square with one right- 
angle side to the T, as shown in Fig. 1, and 
draw with a hard (4H) well-pointed lead pencil 
a line on side a b. Reverse the set square on 
a b as an axis, and if the line drawn and the 
side of the set square coincide (fall into one) 
the angle is a right angle, while a convergence 
will show twice the angle to be corrected, and 
such a set square should not be used until it is 
made true. Likewise investigate the T square 
before using it. 

Figs. 3 and 4. — Plate 3.— Figs. 3 and 4 show the 
different angles possible to be drawn by means 
of both set squares and the T. 

Fig. 2, Plate 1. — TTie protractor is a semi-circular 
instrument made of brass or celluloid, for 
the purpose of measuring the size of an angle. 
Point C represents the center of the semi- 
circle, which is divided by radii into 180 equal 
parts, termed degrees. In measuring an 
angle, such as the angle B C A, place the in- 
strument with its center at the vertex (the inter- 
section of the sides of the angle), and one side 
to coincide with the diameter of the instrument. 
Note the number of divisions on the intervening 
arc, which is 137 (read 137° degrees), which re- 
presents the size of the angle. A degree is fur- 
ther subdivided into 60 parts called minutes ( '), 
which are in turn subdivided into 60 parts 
called seconds ("), 



THE SET OF DRAWING INSTRUMENTS. 



Necessary Tools, Implements and Their application. — Continued. 

sufficiently to have both blades touch paper 
equally to allow an even flow of the ink. The 
leg which carries the center of the compasses 
should have a vertical position so as to avoid 
a tapering of the hole in the paper by its revo- 
lution. 



Pigs. 4 to 7, Plate l.—The very best is none too good. 
A set should contain one pair of compasses, 
Fig. 4, with needle-point center. Pig. 4 D, a lead 
pencil attachment, Pig. 4 B, a ruling-pen for 
circles. Pig. 4 A, one pair of dividers. Pig. 5, 
and one or two straight ruling-pens. Pig. 6, of 
different sizes. Por boilermakers, machinists, 
architectural iron constructors, etc., a set of 
bow instruments is a valuable addition to the 
above. 
Pigs. 7 and 7 a, Plate l.—T/ie lead (6 H) for the 
compasses is bought in sticks of 5 in. in length 
and ^ in. thick. Take a length i in. longer 
than the length of the hole in the attachment to 
receive it. Give the lead the shape shown in 
Pig. 7, which is most conveniently done on a 
piece of emery paper or a fine file ; then take 
off the corners as indicated by the lines G K 
and H I Pig. 7 A. Insert it with a flat side 
towards the center of the compasses and clamp 
it tight with the clampscrew S, Pig. 4 B. 

The main joint near the handle ought to 
move with ease, and one hand should be suffi- 
cient to open or close dividers or compasses 
easily. 

The straight pen and the pen for circular 
ruling must be treated most carefully. Their 
blades are of the same length, not so pointed 
and sharp as to cut the paper, and when filled 
with ink should be entirely free of ink on the 
outside. 

In inking circles, the legs of the compasses 
should be bent at the joints P and O (Fig. 4) 



The correct position of the compasses is 
shown in Pig. 4, where the line M N represents 
the surface of the drawing-paper. A convenient 
arrangement to keep the plates of the course 
for later reference is used in the Chicago High 
Schools. The sheets of paper of 1 1 in. X 17 in. 
are perforated and seamed by laces in a porte- 
folio of 12 in. X 18 in. A rectangle as a bor- 
der line of 10 in. X 15 in. encloses the drawing 
surface which is divided into six equal squares 
5 in. on a side, each to receive one construc- 
tion. See Plate 4. Larger spaces however 
should be used to execute accurately some of 
the constructions, for which the proportional 
sizes may be ascertained from the correspond- 
ing plates. Draw the lines light and carefully 
with Dixon's V H (very hard), Paber 4 H (Sibe- 
rian), or a Hartmuth 6 H (compressed lead) 
pencil, having a fine round point. 

Inking the drawing.— 'Execnte all construc- 
tions in pencil, to admit of corrections, when 
necessary, before inking them. It is also ad- 
visable for the inexperienced to write the re- 
quired text on the drawing in pencil, to distri- 
bute letters and words regularly in the avail- 
able space beneath each drawing, as shown in 
Fig. 1, Plate 4, before writing with Indian ink. 



Plate i. 




Hnimtcln's Comtnictivc Diawlng. 



Plates 2 and 3.— AlpTiabeta.— Several styles of lettering tor 
titles of drawings commonly used, are shown on Plates 
3 and 3. 

Plate 8 —In Fig. 5, ABCDEFQ and H sliow a few samples 
of corners in border lines for elaborate work. 

The following distinctions of inked lines in drawing are made 
to recognize readily all that pertains to problem^ construction 
and reiult. 

The problem line is drawn/ine and uninterrupted. 

The construction line isfl^icand dashed. 

The result, a strong, uninterrupted line. 

Begin inking with construction arcs and circles, then the cir- 
cular problem lines, and then the circular result lines. 

This is done so as to save time, to avoid the change of tool in 
hand, and not to clean and re-set the pen oftener than necessary. 

Construction straight lines are drawn next very fine and 
dashed, corresponding to construction arcs and circles, and 
lastly the 

Result straight line, to correspond to result arcs and circles. 

The inking of a drawing is a recapitulation of each construc- 
tion, and this important work should be executed with great 
care. 



A. postulate is a statement that something can be done, and is 
so evidently true as to require no reasoning to show that it can 
be done. 

An a^^iom is a truth gained by experience, and requiring no 
logical demonstration. 

A theorem is a truth requiring demonstration. 



LINES AND ANGLES. 

A right Une is the shortest distance between two points. 

When the term line alone is used, it indicates a right line. A 
vertical line is the "plumb-line " ; a horizontal Une, one making 
a right angle with the vertical and a line of any other direction, 
is called oblique. 

A curved line or curve changes its direction in every point. 

When two lines lying in one plane, on being produced in either 
direction, do not intersect, these lines are said to be parallel. 

Two lines that intersect, or may be made to intersect, are 
said to form an angle. The point of intersection of these two 
lines, called sides, is the vertex of the angle. 

When two such lines intersect each other, so that all four 
angles formed are equal, we say they are right angles. The 
common vertex, of these four right angles may be assumed to 
be the center of a circle, which by diameters is divided into 360 
equal parts, called degrees ("). Each angle contains % of 360" 
= 90°, which is the right angle. An angle greater than 90° is an 
obtxise angle; an angle smaller than 90° is an cicute angle. When 
the two sides of an angle form a straight line, the angle is called 
a straight migle, and its magnitude is 180°. 

Generally we designate an angle by three letters, for instance, 
. b a c or c a b ; then the middle letter (a) indicates the vertex, 
while the sides are b a and c a. 



Plate 2 



Engineering 'Sct^ipt, (Roman) 
123+5 (vbcclefghi.jkl7n nopqrstuvnvxi^z. ejsoo. 

ABGDEFGHIJh LM^OPQR SnrVWXYZ. 

AF\C^iTtQTVF\AL. 
l2_5-=f^ abcdefgf\ijKlTT\T\opcfKsluVW\ijz. 6/590. 

Shop Skelston. 
abcdeF^hijklmnopciH-stuvwxyz. 

12545 ABCDEFGHIJKLM NOPQRSTUVWXYZ. 87390. 

ABCDEFGHU KLIVINOPQ RSTUVW>{yZ 



ipc|: 



Ha».«(ci»'9 Conslnictlos Drauiinff. 



PLANES AND SURFACES. 

A plane lias two dimensions— length and breadth. 

A surface is the boundary of a body. 

Surfaces bounded by right lines are called polygont. Begular 
polygons have equal sides and equal angles ; they are equilateral 
and equiangular. 



POLYGONS are: 



The triangle, 
" tetragon o 
'* pentagon, 



which has 3 sides, 



quadrilateral. 



heptagon, 

octagon, 

enneagon or nonagon. 



undeoagon, 
dodecagon. 



12 



etc. 



The triangles are: The equilateral triangle which is also 
equiangular; the ieusecles triangle, having two sides equal, and 
the scalene triangle, whose sides are unequal. 

An obtuse and a right-angled triangle have one obtuse and 
one right angle respectively. An acute angled triangle has 
three acute angles. The side or "leg" opposite the right angle 
in a right-angled triangle is called the hypotenuse, the sides or 
legs forming the right angle are the catheti. 

The sum of the squares constructed on the catheti is equivalent 
to the square erected on the hypotenuse. 

The sum of all angles in a triangle is equal to two right 



A line drawn from a vertex of a triangle perpendicular to 
the opposite or produced opposite side is called its altitude or 
height. 



QUADKILATEKALS. 

The regular quadrilateral is the square. (See definition of 
regular polygon.) 

Lines joining the middle of the opposite sides are called 
diameters. 

Lines joining opposite comers are called diagonals. 

In a square the diagonals are equal. 

A quadrilateral whose opposite sides are equal and parallel 
is called a parallelogram, 

A rectangle is a parallelogram whose angles are right angles 
and adjacent sides unequal. 

A rhombus is an equilateral quadrilateral with unequal 
diagonals and equal opposite angles, 

A trapezoid has only two parallel sides. 

A trapez.inn is an entirely irregular quadrilateral. 



Fig. I. Plate 15.— Definition.— A circle is a portion of a plane 
bounded by a uniformly curved line, called the circumference, 
all points of which are equally distant from a fixed point within, 
called the center. 

The distance from the center to any point of the circle is 
called the radius. The connecting line of any two points of the 
circumference is caUed a chord. If the chord is produced to 
any point outside the circle, it is called a secant. The chord 
tiirough the center is called the diameter. Any arbitrary part 
of the circumference is called an arc. The arc that forms the 
fourth part of the circumference is called a quadrant; the sixth 
part a sextant; the eighth part an octant; while half the circum- 
ference is called a semi-circumference. The area comprised by 
two radii and the intervening arc is called sector; the area 
comprised by a chord and the corresponding arc is called a 
segment of a circle. 

In Fig. 1, C D, C B and C A are radii, G H is a chord, E I F is 
a secant, A B is a diameter, G J H an arc, area D C B L D is a 
sector, area H G J H a segment, tract A J U B a semicircle. 

Postulate.— T>T3iVr a circle, if the center and the radius are 
given. 



Plate 3. 



BLOCK, 






EF6HIJKLMNOP0R 

STUYWXYZ. 1234567890. &. 



sijjifloiyuj^ji. 



®^/i^iJjj;iL>7jj^JDj^i;j-F] 



rtr: 





Hmwtein'S'CQi\atntctivc DrawTnQ. 



CONSTRUCTIONS. 



1. — Fig. 1. — Problem. — Tu erect a perpendicular at 
a given poird in a given line, or to bisect a straight 
angle. 

Solution. — Let M N be the given line and A the 
given footpoint of a perpendicular. From A as a 
center and with any radius describe the arc B, C; 
B and C are equidistant from A and are the centers 
of arcs with equal radii greater than B A, which 
intersect at point D. Draw the line D A, which is 
perpendicular to the line M N, in point A. 

2. — Fig. 2. — Problem. — To draw from a given point 
a perpendicular to a given liiie. 

Solution. — With the given point A as a center 
describe an arc intersecting the given line M N in 
two points, B and C. From B and C as centers and 
with equal radii draw arcs intersecting at D. Con- 
nect points A aud D by the line A D, which is per- 
pendicular to M N. 

3.— Figs. 3, 4, 5 and 6. — Problem.— To erect a per- 
pendicular at the end of a given line, M A. 

SoZirfioft.— Take any point C outside of M A as a 
center, and with a radius C A describe an arc in- 



tersecting MA at D. Draw the diameter D C B. 
A line drawn through B and A is the perpen- 
dicular to M A. 

4. — Fig. 4. — Solution. — From A as a center and any 
radius describe the arc B N, at which make 
B C = A B and pass through points B and C 
the line B C indefinite; make then C D = C B. 
A line drawn through A and D is the required 
perpendicular. 

5. — Fig. 5. — Solution. — Describe from A as a center 
and any radius the arc B C E. Make E C = 
C B = B A, and from E and C as centers and 
with equal radii draw intersecting arcs at D. 
A line drawn through D and A is the required 
perpendicular. 

0. — Fig. 6. — SoZtttioji.— From A towards M lay down 
5 equal units. With A as a center and 3 units 
as a radius draw the arc 3 B indefinite, and 
4 as center and 5 units as radius cut the arc 
at B. A line drawn through B A is the required 
perpendicular. 



Plate 4. 



Fig. 1. 


Fig. 2. 


Fig. 3. 


/ 

1 
1 
1 




B 




/ 

/ 
/ 

c/ 
/\ 

/ \ 


1 

1 






1 






M 5\:- — ■ -■;> 


/CH'<imVen«ointmaal,v'en.^inet:» 
^recrapetpeniicufai. 






Fig. 4. 


Fig. 5. 


Fig. 6. 


1 


! 


"K 






/ / 
/ / 

/ / 




/ 

/ 
1 

1 
SI 


~^-s 




B 


M el 






" 5 ♦ 3 2 1 



H«rt!lein'8 C"nstnict(ve Drawing. 



7. — Fig. 1. — Problem.— To co^istruct a perpendicular 
at or near to the end of a given line. 

Solution. — When M N is the given line, take in 
M N an arbitrary point, A as a center and a radius 
longer than A N ; describe are G E D. From an 
other point, B, near N, with any radius, draw arcs 
intersecting arc CED at C and D. Connecting 
points C and D by a line we have the required per- 
pendicular. 

DIVISION OF LINES. 

8.— Fig. 3.— Problem.— To bisect a line. 

Solutio)i. — When A B is the given line, use A as 
center, and with a radius greater than i A B draw 
the arc D C E. With the same radius and center 
B draw an arc to intersect the arc D C E in points 
D and E, which are connected by the line D E. The 
line D E will not alone cut the line A B into two 
equal parts, but will also be a perpendicular to 
AB. 

Figs. S, 4, 5 and 6. — Problem.— I'o cut a given line 
into any number of e(iual or proportional parts. 

9. — Fig. 3. — Problem.— .4 line A B shall be divided 
into 7 equal parts. 

Solution. — Draw the line B N at about 35° to A B. 
Lay thereon, starting from B, seven times a unit 
and connect points 7 and A by line 7 A. Parallel 
with line 7 A, draw lines from each division point, 
6, 6, 4, etc., which will divide A B into the required 
number of 7 equal parts. A C is | of A B. 



Bemark. — Parallel lines are drawn with the set 
and T square combined. Adjust the longest side of 
the set square to coincide with the line to which 
we intend to draw parallels, and place the T to one 
of the right-angle sides of the set square. Keep T 
firmly in this position and slide the set square 
along its edge in the required direction and draw 
the parallels. 

10. — Fig. 4. — Problem.— 2'o cut a given line in two 
proportional parts, «s 3:8. 

Solution. — Draw the line B N, and from B lay 
down a division of 3-J-8 equal parts. Connect 
points A and 11 by the line A 11, and draw parallel 
to it 3 C. C B is i\ and C A Sj of A B. 

11.- -Fig. 5. — Problem. — To cut a given line into three 
proportional parts, us^•.^■.li. 

Solution. — Draw the line B N, and from B lay 
down a division of 7-1-3 J-)-li equal parts. Connect 
point 12i with A and draw parallel to 12J A, the 
lines lOJ C and 7 D. A C is then If, C D 3i, and 
D B, 7 parts of line A B. 

12.— Fig. 6. — Problem. — To cut a given line into any 
number of equal parts by a scale. 

Solution. — Draw a rectangle A 14 N, and divide 
A N by horizontals into any number of equal 
parts, and number them 0, 1, 2, 3, 4, 5, etc. When 
the line A B is to be divided into 9 equal parts, 
take the line to be divided as a radius and A as 
center; describe an arc to intersect line 9 at point 
B 9, which connect with A by line B 9 A. The 
line B 9 A is divided into 9 equal parts by the 
horizontals. 













P/atc 5. 


Fig. 1. 

V ^ 
/ \ 
" — ''''/ i' 


/ 

/' 
/ 
/ 


Fig. 2. 

\ 

\ 
\ 
\ 

--•tc 




Fig. 3. 


\ \ \ V \ \ >'' 
\ \ \ \ \ V^^ 


-' 


1 
( 

1 


A . ; B 
\ ; 

' / 

\ / 

\ / 

> . 
'' E ^ 


A" 


Fig. 4. 

A C B 


Fig. 5. 




B 


Fig. 6. 




"^ i 


''!; 


\v 1 


\\ 


\ \ 






■w 


\ N. 


\ V 


\ \ .' 


\ \'q 3 




^^-0^11 






« 















Hanalein's Constructive DraKlng. 



SOLUTION OP ANGLES. 

13, — Fig. 1. — Problem. — To construct an angle equal 
to a given one. 

Solution. — Angle C A B is the given angle. When 
the vertex O and one side O N of the angle to be 
constructed are given, describe with O as a center 
and A C as a radius the arc E D, and from D as a 
center with the radius B C the arc at E ; draw the 
line E O. Angle C A B = angle E O D. 

14.— Fig. 2.— Problem — To bisect an angle. 

Solution. — Let B A C be the given angle. With 
A as a center and a radius A B draw the arc B C. 
B and C are the centers for arcs with equal radii, 
intersecting at D; draw line from A through D, 
which divides B A C into two equal parts. 

15.— Fig. 3.— Problem.— To trisect a right angle. 

Solution. — From vertex A, with the radius A B, 
draw the arc B C. With B as center and the same 
radius draw the arc A E, and from C the arc A D; 
draw lines through D A and E A. Angle B A D = 
D A E = E A C. 

16.— Fig. 4.— Problem.— To trisect any angle. 

Solution.— Ijei C A B be the angle to be 
trisected. Describe with A as center a semi- 
circle BCD, which intersects the prolonged 



side B A of tbe angle at D ; draw from C an arbit- 
rary line OEM and make E F = E A, and draw 
F G C; then make G H = G A and draw H I C. An 
additional operation will not be necessary, as the 
lines will fall so close together as to almost coin- 
cide, and it is angle C H B which equals i C A B. 
This construction is convenient for angles up to 
90"; and in case of the trisection of an obtuse angle 
we bisect first and then trisect, so that the double 
third of the bisected angle is equal to the third of 
the given obtuse angle. 

SOLUTION OP TRIANGLES. 

17. — Fig. 5. — Problem. — To consti-uet a triangle when 
the three sides are given. 

Solution.— Liines 1, 2 and 3 are the sides given. 
Lay down line 6 = line 1. From C as center, 
with line 2 as a radius, draw an arc, and with line 
3 as radius and center B another arc, intersecting 
the first arc at D. Draw lines D C and D B ; then 
D C B is the required triangle. 

18.— Fig. 6.— Problem,— To construct a triangle of 
which tioo sides and the included angle are given. 

Solution. — Construct angle D, and from its vertex 
cut off the sides 1 and 3, that is C B and C E, and 
draw line E B; then E B C is the required triangle. 



Fig. 1. 




0} 



Fig. 2. 




Plate 6 
Fig. 3. 




Fig. 4. 



Fig 5. 





Fig. 6. 




Hanstein's Constructive drawing. 



19. — Fig. 1.— Problem. — To construct a triangle of 
which one side (1) and the two adiacent angles D 
and E are given. 

Solution. — Lay off C B equal to line 1; transfer 
the angles D and E on line C B, and prolong the 
sides to intersect at P; then triangle C F B is the 
required triangle. 

20.— Fig. 2. — Problem. — To construct a triangle of 
which one side (1), one adjacent angle D and one 
opposite angle E are given. 

Solution. — Construct C B equal line (1) and angle 
D at C as before; at an arbitrary point M on line 
CM draw angle C M N = E, and parallel with M N 
the line B F. F is the third vertex of the required 
triangle C P B. 

PROPORTIONAL LINES. 

21.— Fig. 3.— Problem.— To construct to three given 
lines a fourth proportional. 

Solution. — Let 1, 2, and 3 be the given lines. Lay 
down an angle M A N of about 40", and from A cut 
the segments A 1 = line 1, A 2 = line 2, A 3 = line 
3; draw line 2 1, and with it parallel the line 3 x. 
A X is the required line. 1 : 2 = 3 : A x. 



22. — Fig. 4. — Problem. — To construct to two given 
lines a third proportional. 

Solution. — Let 1 and 2 be the given lines. Lay 
down the angle as before, and from A cut the seg- 
ments A 1 = line 1, A 2 = line 2, A 2' = line 2. 
Draw line 21, and parallel with it 2' x. Ax is the 
required line. 1 : 2 = 2 : A x. 

23. — Pig. 5. — Problem. — To construct a mean pro- 
portional to two given lines. 

Solution. — Let 1 and 2 be the given lines. A B -f 
B C is the sum of the given lines 1 + 2. Find point 
D, the center of A C, and with a radius D A, draw 
the semi-circle A X C. Erect at B a perpendicular 
B X, which is the required line. A B : B X = 
B X : B C. 

24. — Fig. 6. — Problem. — To construct to a given line 
major and minor extreme proportionals. 

Solution. — At point B of the given line A B erect 
a perpendicular B D = i A B, and draw line A D P 
indefinite; with D as center, D B as radius, describe 
semi-circle EBP, and from A as center, A E as 
a radius, draw arc E X. The line A P : A B = 
A B : A X. 



Fig. 1. 



A 



Fig. 2. 




Plate. 7. 
Fig. 3. 



\ s 



_-b "s- - 



Fig. 4. 



\ \ 
\ \ 
\ \ 
i ^v. 

2 I 



Fig. 6. 




HarxsUiri's. Constructnvc Drmoing, 



POLYGONS. 

25.— Fig. 1.— Problem.— To construct a reoular triangle on "■ 
given bane. 

Solution.— A B is the given base With A and B as centers and 
A B as radius draw arcs intersecting at C. Draw the lines C A 
and C B. A C B Is the required regular triangle. 
36.— Fig. 1.— Problem.— To eon-strttct a regular hexagon on a 
given base. 

Solution.— het A B be the given base. Construct on this a 
regular (or equilateral) triangle. The vertex C is the center, 
and C A = C B the radius of a circle, in which a regular hesa- 
gon A B D E F Q, v;ith A 6 as side, can be inscribed. 

Corollary. — A regular hexagon may be divided into six equal 
equilateral triangles, the common vertices of which Ue in the 
center of it. 

37.- Fia. 2.— Problem.— To construct a regular heptagon at a 
given base. 

Solution,— Draw with the given base A B the equilateral tri- 
angle A 6 B. as in the previous construction. From center D of 
A B draw the line D 6 13 perpendicular to A B. Divide 6 A into 
six equal parts. These parts transfer on line 6-1.2 and number 
them, 7, 8, 9, 10, 11 and 13. Point 7 is the center, and 7 A the 
radius of a circle, in which the regular heptagon A B C D E F G, 
with A B as aide, can be inscribed. 

28.— Fig. 2.— Problem.— To construct a regular polygon loith 
more than 6 sides. 

Solution.— With points 7, 8, 9, 10, 11 and 13 as centers, and 7 A, 
8 A, 9 A, 10 A, 11 A and 13 A, respectively as radii, draw circles 
in which the line A B rs repeated chord will form the regular 
heptagon, octagon, enneagon, decagon, undecagon and dode- 
cagon. 

Bemarfc.- Regular polygons with greater number of sides are 
rarely used in practice, and are therefore omitted here. 
29.— Fig. S.— Problem.— To construct a square at a given base. 

Solution.— \je.t A B be the given base. Draw at A and B per- 
pendiculars with set and T square, and make A C = A B, and 
with T square draw CD. A O D B is the required square. 
30.— Fig. S —Problem.— To construct a regular octagon at a 
given base. 

Solution — In the bisecting point H of the given base A B 
erect a perpendicular. H F, at which make H E = A H and E F 
= E A. F Is the center and F A the radius of a circle, in which 
draw A B eight times, as repeated chord, to complete the re- 
quired octagon ABGHIJKL. 



31.— Fig. 4.— Problem,— To cointrifct a regular pentagon at a 
given base. 
Solution.— Let A B be the given base; produce it towards N. 
Erect at B a perpendicular, B D = A B. Bisect A B by point C; 
with C as center and C D as radius draw arc D E. With A and 
B as centers and A E as radius draw arcs to Intersect at F. 
With F and A as centers draw arcs intersecting at G; and from 
F and B as centers, with the same radius A B, draw arcs inter- 
secting at H. Connecting B H, H F, F G and G A by lines we 
complete the required pentagon A B H F G. 

32.— Fig. 4.— Problem.— To constn«;t a regular decagon at a 
given base. 

Solution.— Let A B be the given base. Follow the construc- 
tion of the pentagon until the position of i)oint F is found; this 
is the center, and F A the radius of the circle, In which as re- 
peated chord the line A B will complete the required regular 
decagon ABIJKLMNOP. 

33.— Fig. 5.— Problem.- To construct triangles equivalent to a 
given one. 

Solution.— Let A B be the given triangle; draw line M N 
parallel with A B through point C. Locate an arbitrary point 
E or G in line M N, and draw lines E A and B B, and G A and 
G B. Triangle A B B = A C B — A G B. It one side of the tri- 
angle is called the base, a perpendicular drawn from the oppo- 
site vertex to the base, or produced base, is the altitude or 
height of the triangle, as E F, C D and G H. 

Theorem. — Triangles of equal base and altitude are equi- 
valent. 

84.— Fig. 5, A.— Problem.— To construct paraUdograms equiva- 
lent to a given one. 

Solution. — Let A B D C be the given parallelogram, with base 
A B. Draw the line M N parallel with A B, make B F and G H 
= C D, and draw lines E A, F B, G A and H B. The parallelo- 
gram BFBA — CDBA = GHBA. 

In a polygon any right line which passes through two non- 
consecutive vertices of its circumferential angles is called a 
diagonal. 

27ieorem.— Either diagonal divides the parallelogram into two 
equal triangles. 

3B.— Fig. 6. — Problem.- To construct a rectangle equivalent to a 
given triangle. 

Solution.— A B C may be the given triangle, and C F its alti- 
tude. Bisect C F rightangularly by line D E, and erect the per- 
pendiculars B £ f nd AD. A D E B is the required rectangle 



Fig. 1. 





Plate 8. 

Fig. 3. 




Fig 4. 





Fig. 3. a 



Hajisleln's Coiistniclfve Drawing. 



36.— Fig. 1.— Problem.— To construct a rectangle 
equivalent to a given trapezoid. 

Solution. — Let A B C D be the given trapezoid. 
Bisect rightangularly its altitude LM by the line 
I K, which bisects also the sides B A and C D in I 
and K. Perpendicular to I K, through I and K, 
draw F G and E H to intersect the produced B C 
in E and F. F E H G is the required rectangle, 
equivalent to the trapezoid A B C D. 
37. — Fig. 3.— Problem.— T/ie side of a square is given: 
to construct the sides of squares that are twice, 
three times, four times, etc., as great as the square 
over the given line. 
Solution. — Construct a right angle B A 1 ; make 
B A and A 1 equal to the given side of the square; 
then lay off successively A2 = B1,A3 = B2, A4 
= B 3, etc. A 2, A3, A 4, etc., are the sides of 
squares that are respectively twice, three times, 
four times, etc., the area of the square over A 1. 
38. — Fig. 3.— Problem.— To construct a triangle equi- 
valent to a given irregular pentagon. 
Solution. — Let A B C D E be the irregular penta- 
gon. By the diagnols A and C E divide it into 
three triangles A B C, C A E and C D E. Produce 
the base A E to the left and right indefinitely, and 
parallel to C A draw the line B F ; connect C with 
F ; then draw D G parallel with C E and connect C 
with G. The sum of the triangles C F A + C A E 
+ C E G is equal to the triangle CFG, which is 
equivalent to the irregular pentagon ABODE. 



39. — Fig. 4.— Problem— 2'o co7istruct a square equi- 
valent to a given triangle. 

Solution. — Let CFG, Fig. 3, be the given tri- 
angle. Construct a mean proportional between 
half the base F G and altitude C H, as shown in 
Fig. 5, Plate 7, by making I K -^ i F G, and K L = 
C H. The sum I K + K L is the diameter of the 
semicircle I N L. Erect at K a perpendicular, which 
is intersected by the circle in N. N K is the side 
of the required square, and N O P K is the square, 
which is equivalent to the triangle CFG and the 
irregular pentagon A B C D E. 

40.— Figs. 5 and 6.— Problem.- To tramform anir- 
regular heptagon into an equivalent triangle and 
square. 

Solution.— Let ABCDEFG be the irregular 
heptagon. Draw line C A, and parallel to it B N; 
connect N and C by line N C. Triangle C N A = 
C B A. Treat the triangle E F G in a similar way, 
and you have transformed the heptagon into the 
irregular pentagon N C D E M. Proceed as in Fig. 
3, and transform the pentagon into the triangle 
D H I; transform this into the square P Q R O, 
Fig. 6, which then is equivalent to the given hep- 
tagon ABCDEFG. 



Plate 9. 







Fig. 1. 






Fig 2. 




C 


f;-. 3. 


; 


6 


L C 


E 


B 


^ 




Z' 


\/' ' ^ 


\ / " 

V \ 


/->.-. 


-.-.^ 


\ 




\ \ 


A MHO 


A. 1 




^3 °". M 


F 


A H 


E G 






Fig. 4. 






R^'. 5. 






Fi^-. 6. 


/ 

/ 

1 
1 
1 
1 
1 


N 




\ 
\ 
\ 

; 


# 


\/A 


/ V 


/ 

/ 
/ 
1 

1 
1 

L 


P 






Q 




! 

: M 




A 


/M^y 






N 


\ 
\ 

\ 


K 1 


P L 


H " L 


M C 


' 


° i 








•i; 












V 





Wan3(eiri's Ci>ns(nM^(u"c Draiofno. 



TO TRANSPEK POLYGONS. 

41. — Figs. 1 and 2.— Problem.— Tb coiistruct a poly- 
gon equal to a given one by triangles. 
Solution. — Let Pig. 1 be the given polygon. 
Divide it by diagonals into triangles. Draw line 
A' B' parallel and equal to A B. Upon this con- 
struct triangle A' B' C equal to triangle ABC. 
Lay off the remaining triangles of Fig. 1 in the 
same order and position, starting from side B C; 
then polygon Pig. 2 is the required one. 

42. — Figs. 3 and 4.— Problem.— 2'o construct a poly- 
gon equal to a given one by sectors. 
Solution. — Polygon Pig. 3 is given. From center 
O with any radius describe circle C B D. Draw 
from center O a radius to each vertex of the poly- 
gon to intersect with the circle. Locate center O', 
Fig. 4, and with radius O' D' equal O D describe 
the circle D' B' C, and draw O' D' parallel to 
OD; make arcs D' B' = D B, B' C = B C, etc., and 
pass lines through points D', A', C, etc.; further 
make O' E' = O E, O' A' = O A, O' F' = O F, etc., 
and by connecting points E' A' F', etc., complete 
the required polygon. 



43. — Figs. 5 and 6. — Problem. — To construct a poly- 
gon equal to a given one, by co-ordinates. 

liemark. — In the plane of drawing a convenient 
line is drawn (horizontal), called the axis of ab- 
scissae; the position of the diiJerent vertices of the 
given figure is determined by perpendiculars (or- 
dinates), from these vertices to the axis of abscissae. 
Take any convenient point, A, on this axis and 
draw a perpendicular to it, M N. This line is 
called the axis of ordinates, and reckoned from 
this point A (called the origin) the segments deter- 
mined by the foot-points of the ordinates are called 
abscissae. Abscissae and ordinates together are 
called co-ordinates. 

Solution. — Fig. 5 is the given polygon. Through 
any vertex (origin) draw a horizontal, A R, then 
M N becomes the axis of abscissae. Draw the 
ordinates from each vertex or principal point for 
transmission perpendicular to A R, the axis of 
ordinates. Next draw A R', Pig. 6, and lay oflf 
A B, AC, AD, etc., = A B, A C, A D, etc., of 
Fig. 5. Erect the perpendiculars A A', B B', C C, 
C", C", etc., and make A A', B B', C C, O C", 
C C", etc., equal to the corresponding perpendi- 
culars in Fig. 5. Connect A and A', A' and B', 
describe with radius C'C, center C, arc B' C", etc., 
and complete the required polygon, Pig. 6. 



Fig. 1. 




Fig. 2. 




Plate 10. 

Fiff. S. 




--^A« 



Fig. 3. 



Fig. 6. 






Han.ftein^i Conftnicttvc Drawing. 



44.— Figs. 1 and 2.— Problem — To construct a poly- 
gon equal to a given one, radAating in a circle. 

Solution. — Let A E D G, etc., be the given poly- 
gon. Describe withi AD, A E, etc., as radii and A 
as center tlie circles C D, E F G, etc., and make 
P' E' = F E, F' G' = P G, etc. Connect D' and E', 
D' and G', etc., and complete the required polygon. 
Fig. 2 shows the construction applied to other 
polygons. 

EemarJc. — This construction is used conveniently 
to draw a rosette in which an ornamental unit 
occupies a sector division of a circle. 

45.— Figs. 3 and 4.— Problem — To coiistruct sym- 
metric polygons or outlines. 
Solution. — LetL M, etc., be the given outline as 
a profile of the base of a column. Draw the hori- 
zontals L L', M M', etc., and the axis of symmetry 
R N. Make A L' = A L, B M' = B M, D O' = D O. 



etc. Connect L' and M', etc., and complete the re- 
quired symmetric profile of the base of the column. 
46. — Figs. 5 and 6. — Problem.— To construct an ir- 
regular outline equal to a given oiw. 
Solution. — Let B A C be the given outline. Cover 
this with a series of equal small squares and con- 
struct in Fig. 6 the same number of equal squares 
arranged as in Fig. 5, and transfer the points of 
intersections of the irregular outline with the sides 
of the squares ; make M' A' = M A of Fig 5, and 
M' B' = M B, etc. Connect B' A' C by a free-hand 
line and complete the required irregular outline, 
Fig. 6. 



Fig. 1. 



Fig. 2. 




Plate 11. 
Fig. 5. 




I [__!_ 



Fig. 3. 




Fig. 4. 




Hanstein's Constructive Drawing. 



TO SEDUCE OB ENLARGE POLYGONS IN ODTLIHE OB AREA. 

47.— Fias. 1, 2 and 3.— Problem.— To construct a polygon efm 
ilar to a given one of 4 its circumference. 

Solution.— Let D A B C, etc., Fig. 1, be the given polygon. 
Construct the Scale Fig. 2. A perpendicular O 7, longer than 
the longest side of the given polygon, is divided into 7 equal 
parts ; draw a horizontal line O N of an arbitrary length and 
connect points 7 and 4 with N by the lines 7 N and 4 N. O 4 is 
A, — 4 7 is a of the line 7. All lines between O N and 7 N 
and parallel to O 7 are divided by 4 Sf and 7 N in the same pro- 
portion. To obtain the length of A' B', Fig. 8, place line A B in 
the scale as indicated by line A B' B, of which A B' is ^ of line 
A B. Transfer the remaining sldei of the polygon by parallels 
and find of each the proportionate length in the scale Fig. 2, as 
shown by line A B; D' A' B' C, etc., is the required polygon. 

48.— FiQS. 1, 4 and 5.— Problem.— To construct a polygon Sim 
ilar to a given one, having * its area. 
SoJution.— Let D ABC, etc., Fig. 1, be the given polygon. On 
a horizontal line O 4 lay down a division of 7-{-.4 squal parts and 
malce O 4 the diameter of a serai-circle O M 4. Erect at point 7 
the perpendicular 7 M and draw lines M O and M 4. Then make 
line M B' equal to A B of the given polygon and draw B' B" 
parallel to O 4; A" B" (Fig. 5) — M B" in the scale Fig. 4. In 
relation to the side A B of the given polygpn. A" B" is the side 
of a polygon, whose area is } of the given one. Treat the re- 
maining sides of the polygon similar to the side A B and com- 
plete the required polygon D" A" B" C", etc. 



49.— Fig. 6.— Problem.— To construct simitar polygons which 
have I the circumference and | the area of a given one. 

Solution for circumference reduction.— Let D C B A E, etc. , be 
the given polygon. From any point O therein draw radii to the 
vertices D C B A E, etc., and divide any one radius (O D) into 5 
equal parts. Parallel to D C from point 3 draw D'C, with 
B, 0' B', etc., and D' C B' A B, etc., is the required polygon. 

Solution for area reduction.— Hake radius O D the diameter 
of the semi-circle O N D and erect at division point 3 the per- 
pendicular 3 N and draw N O. Make O D" =« O N and proceed 
as before in drawing D" C" parallel with D C, C" B" with C B, 
etc. D" C" B" A" E", etc., is the required polygon. 

50 — Figs. 7, 8and9.— Problem.— To reduce any irregular out- 
line in proportion 8 :5. 

Solution.— Let Q H I K be the given irregular outline . Cover 
the given outline by a net of equal squares, the sides of which 
we reduce by the scale, Fig. 8, to A' B' = % of A B. Draw with 
A' B' as unit the same number of squares as in Fig. 7. Transfer 
the points of intersection of the irregular outline with the sides 
of the squares, in reducing their distances from the vertices of 
the squares by scale Fig. 8, and transfer into Fig. 9. Connect 
these points by a free-hand line, which is the required reduction 
of the irregular outUne. 

Treat the surface reduction, Fig. 11, with the assistance of 
the scale Fig. 10 in a similar way, and we obtain the reduction 
in area. 



Plate 12. 



Fig.l 



FTg.3. 



Fig. 5. 



Fig. 4. 




Fig. 6. 




Fig. 







Fig.ll. 



Fig. S. 




Fi 


?■ 


9 




r 


■^ 




^'f 








■( 


6' 














\ 
















1 












V-41 1 



/■n-E/ 


jAE" 


"^ 




^'i' 


/ 


I 


\ 


-4 


FF 






1 




^~ 





HffTMfcfn's Cmr.'rtrHoMrj'- Dtaieinfc 



51.— Figs. 1, 2 and 3.— Problem.— ro construct a 
polygon siinilar to a given one, and of | its cir- 
cumference. ( Transfer hy triangles. ) 

Solution. — Let A B D C, etc., be the given poly- 
gon. Construct the linear scale in proportion 2:3 
Fig. 2 similar to Fig. 2, Plate 12, and divide the 
given polygon by diagonals into triangles. Line 
A B' in the scale (Fig. 2) = A' B' of the polygon 
Fig. 3, whose circumference contains 3 units to 2 of 
the given polygon. Transfer and complete by tri- 
angles the required polygon A' B' D' C, etc., Fig. 3. 

53.— Figs, l, 4 and 5.— Problem.— To construct a 
polygon similar to a given oiie, which contains S to 
each S square units of the given polygon. 

Solution. — In the scale Fig. 4 the diameter of the 
semicircle consists of 3 + 3 equal parts; erect 2M. 
Draw M O and M 3. Make M B' Fig. 4 = A B of 
the given polygon and draw B' B" parallel to O 3, 
A" B" = M B" of the polygon. Fig. 5, whose area 
has 3 square units to 2 of the given polygon. 

Transfer and complete by triangles the required 
polygon A" B" D" C", etc., Fig. 5. 



53.— Fig. 6.— Problem.— To construct a scale of da 
cimal division. 

Remark. — Small subdivisions of a unit which we 
cannot accurately perform with the dividers are 
constructed in Figs. 6 and 6 A. 

Solution. — Let line R 7=8 centimeters, RO=RF 
= 1 cm. The decimal subdivision (millimeter, 
mm) is obtained by dividing O N into 5 equal parts 
by the horizontals in points A, B, c and d. Bisect 
F N and draw lines 5 R and 5 O; line a 1= I'o, 
B 2 = i?o, c 3 = i'^,, etc., of O R, or 1, 2, 3, etc., mm. 
the required division. 

54.— Fig. 6 A.— Problem.— To divide a centimeter 
into 100 equal parts. 

Solution. — Upon a straight line A 7 lay off eight 
units (cm) and construct squares on these distances. 
Let first square be A B N O. Divide sides A B and 
B N into ten equal parts (mm). Draw horizontals 
through division points on A B. R, being the 
first point of division from N to B, is connected 
with O, and through the other points parallels to 
R O are drawn between B N and A O. These 
parallels subdivide the millimeter (mm) into tenths- 



Plate 13. 



Fig. 1. 



Fig. 2. 



Fig. 3. 



Fig. 4. 




STJj^t^X . 




6A4 














V .\ 














-M-A- 


6 












f — H 


' 













Fig. 6. A {^ai^tctioii g^xj,f« in 31tx:t*x 




HniKsfein's Constnictirc DraniC(7. 



55. — Fig. 1. — Problem. — To construct a scale in which 
an inch is divided into 6Iiihs. 

Solution. — Let A 2= 3 inches. Divide BN and 
B A into 8 equal parts each and complete the scale 
in the manner explained in Problem 54, Plate 13, 
Fig. 6 A. Line R O divides line R N = i in. into 
8 equal parts, hence into 64ths. Example : Take 
from this scale a line of 1|J inch ( |J = l + gV). 
From O to 4 = f in.; follow the oblique line 
upward to the 5th horizontal point, N. Line N A = 
I in., A B = /:[ in. and B M = 1 inch and line N M = 
IJI in., as required. 

REDUCTION SCALES. 

56.— Figs. 2 and 8.— Problem To construct a de- 
cimal rediiction scale and draw by co-oi'dinates 
a polygon whose equations are indicated at tables 
A and B, Fig. 2 A. 



Remark. — To draw the scale and polygon in con- 
venient proportion let the unit O A = 2i in., which 
may represent 100 feet. 

Solution. — Let O A be the unit to represent 100 ft. 
in the decimal reduction scale and let A 300 = 3 
such units. Divide O A, A B and B N into 10 equal 
parts, draw horizontals from 9, 8, 7, etc., and the 
oblique parallels with R O from division points 
10, 20, 30, etc., and we have the required decimal 
scale. Example: Take from this scale a line to 
represent 178 feet. Begin at point O, pass to the 
left to 70, then upward the oblique line to the third 
horizontal point R. Line r a = 70 ft. a b = 3 ft. 
and B S = 100 ft., and ra + ab + bs = 173 feet. 

The polygon, Fig. 3, is constructed with this 
scale. 

Bemark.—lt the scale. Fig. 2, is used as a reduc- 
tion scale in which O A represents 1 ft. , we shall 
have to divide O A into 12 equal parts (inches), etc., 
and the scale will represent j', of actual dimension. 



Plate 14. 



Fig. 1. 



SucA -bivliei) i,n,64'-" 




Fig. 2. A 



Sci^e^ A 



A B 
A C 
A D 
A E 
A F 
AG 
A H 
A I 
AJ 

A K _ 
A L = 



5. Ft 
27 ■ 

4 8,5- 

5 8. » 
1 5.5" 
10 9.' 
II?.- 
128 " 
13 9. » 
I 52 
I 5 3. 



%^U^ B 



A M = 2 2 K 



BR 


„ 


3 6 5.Ft 


C 


— 


51. » 


uiy 


as 


7 5 " 


V v 


_ 


2'5 5 " 


PF' 





3 2 


(iCV 


„ 


6 3. " 


HH' 


-_ 


5 0.5 .' 


1 r 


_ 


32 » 


^J J 


:^ 


50.5 " 


KK' 


_ 


3 1.5 • 


1 1 ' 





109.5 • 


MM' 


= 


9 • 



Fig. 2. Sl/abu<txo'n SAixlAinS/ui:, i"eoo 



11^; 




Uanaieln's Constructive Drawing. 



DIVISION OF CIEOLES. 

57. — Fig 2.— Problem.— Tu inscribe a regular trianale, hexagon 
and dodecLKjon in a t/iue;i circle. 

Solution.— Let A B F D be the given circle. Describe with 
point A as a center and radius .V C tUe arc BCD and draw line 
B D, which is the side of the required regular inscribed triangle. 

Heragon.— Line B A = radius, B C = the side of the required 
regular inscribed hexagon. 

Dodecagon.— Bisect the arc B A by point E; draw B B, which 
is the side of the required regular inscribed dodecagon. 

68.— Fia. 3. — Problem. — To insci'ihe in a given circle, C, a 
square, octagon and a regular polygon of 16 sides 



Solution.- 
DG. Dra« 



-Construct two perpendicular diameters, A B and 
D B, which is the aide of the inscribed square. 



Ocfafloii.— Bisect the quadrant D A (in E) and draw D E, which 
is the side of the required regular inscribed octagon. 

The regular polygon of 10 sWe«.— Bisect the arc D E by F, and 
draw D F, which is the side of the required regular inscribed 
polygon of 16 sides. 

69.— Fig. 4.— Problem.— To inscribe a regular penta{/nn atid 
decagon in a given circle. 

Solution.— Draw two perpendicular diameters, A B and E I, 
in the given circle C. Bisect radius C B at point D, and with 
D E as radius, D as center, describe arc E P and draw line E G 
= E F, which is the side of the required regular inscribed pen- 
tagon. 

Decaffon.— Bisect the arc E G by point H and draw E H, which 
is the side of the required regular inscribed decagon. 



60.— Fig. 5.— Problem.— To inscribe a regular lieptagon and a 
regular polygon of 14 sides in a given circle. 

Solution.— Draw a radius, A C. With point A as center and 
A C as radius describe arc BCD and draw BED. HF — FD — 
D E = the side of the regular heptagon in the given circle. 

Bisect arc P H by point G and draw H G, which is the side of 
the regular polygon of 14 sides in the circle. 



RECTIFICATION OP ARCS. 

61,— Pig 6.— Problem.— To rectify a gioen arc. 

Solution. —het A B, corresponding to angle A O B, be the 
given arc. Bisect angle A O B by O N and bisect also angle 
A O N by O N'. Erect B D perpendicular to O B at B, D' D per- 
pendicular to O N at D, D' G perpendicular to O N' at D', and 
draw arc D' H with radius D' and center O. Divide H G into 
three equal parts, and from the first division point J, near H, 
drop J L, a perpendicular to B, then J L — arc BOA. The 
approximation is very close as long as the given angle does not 
exceed CO"; but for greater angles, the half of them may be 
rectified. 

From the rectified arc we can find the area of the correspond- 
ing sector: construct a triangle with the rectified arc J L as 
base and with the radius of the circle as the altitude; this tri- 
angle has the same area as the sector in question. — To transform 
a circle into an equivalent square, we may rectify the arc of 
45°, construct a triangle that has for a base 8 times the length 
of this arc, and for the altitude the radius. Transform this 
triangle into a square, then this square will be equal to the area 
of the circle. — In order to find the length of the circumference 
of a circle we would rectify the arc of 45'^ and multiply this 
length by 8 



F'ff- 1- 




Fig. 4. 




Fi.c:. 2. 





Fig. 5. 



Plate 15 

Fig. 3. 




Pig- 6 




Hansteln's Constructive Drawing. 



62. — Fig. 1. — Problem.— To construct a line equal to 
the semi-circumference of a given circle. 

Solution. — In the given circle C draw two perpen- 
dicular diameters, A B and F G, and, at G, the in- 
definite line E H perpendicular to F G. With A as 
center and A C as radius describe arc C D and draw 
line ODE. Make E 3 = 3 A C and draw F 3 = G 
H, which is equal to the semi-circumference of the 
circle C. Calculation gives — 

F 3 = 8.14153 times radius; 
error = 0.00006 of semi-circumference. 
Denoting the ratio of the circumference to the 
diameter of a circle by the letter tt, then this ratio 
has been more accurately found to be 

IT = 3.1415926; 
for common usage it suffices to take for it — 
■7T = V = 3. 14:28, with an error = 0.001. 
Among the many approximative methods to rec- 
tify a circle, the above method has the advantage 
that it can be performed with one opening of the 
compasses. 

TANGENTS. 

63.— Fig. 2.— Problem. — To construct a tangent at a 
given point of a circle. 

Definition. — A tangent is a line touching the cir- 
cumference of a circle in one point only, the point 
of contact, and is a perpendicular to a radius, 
drawn to the point of contact. 

Solution. — Let C be the given circle and A the 
point of contact. Draw the radius C A, and per- 
pendicular to it, at point A, the line M N, which is 
the required tangent. 



64. — Fig. 3. — Problem. — From a given point outside 
a circle to draw tangents to this circle. 

Solution. — Let C be the given circle and A the 
outside point. Draw A C, and on A C as a dia- 
meter describe a circle, center B; this circle B in- 
tersects circle G at points O and P; then lines A O 
and A P are tangents to circle C. 

65.— Fig. 4. — Problem. — To construct common ex- 
terior tangents to two given circles. 
Solution. — Let C and A be the given circles. Draw 
line C A and upon this as diameter, circle C A; 
with the difference C F, of the radii D A and C E 
draw are H P I from center C, intersecting circle B 
at points H and I. Draw radius C O through H, 
and OP through I. Radii AO' and AP' are parallel 
to CO and CP respectively. O'O and P' P are 
the points of contact of the common tangents. 

66. — Fig. 5. — Problem. — To construct common interior 
tangents to two circles. 

Solution.— Follow the previous construction and 
describe the circle C F A G. With the sum of the 
radii of both circles AD + CI = CE draw arc 
PEG; also the lines C P and its parallel radius 
A P', and C G and its parallel A O'. The inter- 
sections O and O', P and P' are the points of con- 
tact of the required tangents P P' and O O'. 



Fig. 1. 




Plate 16. 
Fig. 2. 




Fig 3. 






ihinaiiirCA Cunsiructivt Draidno. 



TANGENTIAL CIRCLES. 

67.— Fig. 1.— Problem — To construct circles, D and 
H, that touch a given line, M N, and a given circle 
in point A. 

Solution. — Draw line H C A B through center C 
and the given point of contact A; at A erect a per- 
pendicular to H B, intersecting M N in point E. 
With E as center, E A as radius, draw the semi- 
circle FAG and erect at P and G perpendiculars 
to M N, to obtain on line H B the intersections H 
and D, which are the centers, and H A and D A the 
radii respectively of the required tangential circles. 

68.— Pig. 2.— Problem.— To construct a circle of a 
given radius that touches a given circle and a given 
line. 

Solution. — Let C be the given circle. M N the 
given line, and R S the given radius of the re- 
quired circle. 

Draw R' O parallel to M N at a distance R' S' 
equal to R S. With C as center and radius equal 
to the sum of radii of given and required circles, 
as radius, cut R' O in A. This is the center of the 
required circle. 

69.— Fig. 3.— Problem. — Within a given triangle to 
inscribe a circle. 

Solution. — Let A B C be the given triangle. Bisect 
two angles, A and C, by A D and C D, which inter- 
sect in D. Draw the perpendicular D E, which is 
the radius, and D is the center for the inscribed 
circle. 



70. — Fig. 4.— Problem.— To circumscribe about a given 
triangle a circle. 

Solution. — Let B A D be the given triangle. Bisect 
two of the sides by perpendiculars, which intersect 
in the center of the required circle. 

71.— Fig. 5.— Problem.— To connect any number of 
points by a regular curve. 

Solution. — Let ABODE, etc., be the given points. 
Draw lines A B, B C, C D, etc., and bisect each by 
a perpendicular. Take an arbitrary point G at the 
bisection line G N as a center, and with G A as a 
radius draw the arc A B; draw then B G H, a line 
to intersect the bisecting perpendicular of B C in 
H, the center, and H B the radius of the arc B C; 
I is ths center, radius I C for arc D, etc. Com- 
plete the required curve to point P. 

72. — Fig. 6.— Problem, — To construct a curve to the 
base of an Ionic column. 

Solution. — Let A D and D H be the given di- 
mensions. Trisect A D and draw in B (1st 3d) 
a perpendicular, K B E; B A is the radius 
and B the center of quadrant A K. Make B E, and 
EP = BN = JBA and draw F E N L; E is the 
center, E K the radius for arc K L. Erect at H a 
perpendicular, H G, indefinite, at which make 
H I = L F, and draw and bisect F I by the perpen- 
dicular M J, which produced will give the intersec- 
tion point G; draw line G F O. With F as center, 
F L as radius, describe arc L O; with G as center, 
G O as radius, the arc O H. 



Plate 17. 




Fig. 2. 







'i 



i.f-- / 



■■]: 



Hanstttii'a Constructice Drawing- 



J 



73.— Fig. 1.— Problem. — To construct three tangential 
circles token their radii are given. 
Solution. — Let A, B and C be the given radii. 
Draw line G F E = A + B. Describe circle G with 
radius G F = A, .ind circle E with radius E F = B. 
With G as center, and A + C as radius, E as cen- 
ter, B + C as radius, draw arcs intersecting at H. 
H I is the radius and H the center for the third re- 
quired tangential circle. 

74.— Pig. 3.— Problem. — To construct three tangential 
circles when the three centers are given. 
Solution.— Liet A B be the given centers. Con- 
struct the triangle ABC. Make C D = C B, A E 
= A B, and bisect D E in point P. Describe the 
required circles from points C, A and B, as centers, 
with radii C F, A P and B H. 

75. — Figs. 3 and 4. — Problem.— To constiitct tangen- 
tial circles within a given angle. 
Solution. — Let A B C be the given angle, which is 
bisected by A D. Draw a perpendicular line D C 
at an arbitrary point D to form angle D C A, which 
is bisected by C E. The intersection of A D and 
C E is point E; with E as center, and with the radius 
E D describe the tangential circle D P. Perpen- 
dicular to A D, at point F, draw P G, and parallel 
with C E, G H. H is the center, H F the radius for 
the next circle, etc., etc. 



76.— Pig. 4.— Solution 2.— Bisect the angle BAG 
by A D and at an arbitrary point, E, erect the 
perpendicular E P. Make P H = E F and draw 
perpendicular to A B at H, H I. I is the cen- 
ter, I E the radius of the circle E H J. Re- 
peat this construction by making L M = L J, 
etc., etc. 

77. — Figs. 5 and 6.— Problem. — To construct any 
number of equal tangential circles within a given 
circle. 

Solution. — Let C A B D be the given circle. Divide 
the circle into double the number of equal parts as 
you intend to draw circles therein; for 3 circles into 
6, for 5 circles into 10 equal parts. 

Construct at an intersection of diameter and cir- 
cumference point A a tangent to intersect the pro- 
duced adjoining diameter in E. Bisect angle AEG 
by E F ; F is the center, P A the radius for one re- 
quired circle. With center G of the given circle 
and radius G F draw circle P I H, to obtain I and 
H, the centers of the required remaining tangential 
circles. 

Problem Pig. 6 is solved in a similar manner. 



Fig. 1 




Fig. 2. 




Plate 18. 
Fig 3. 




Fig. 4. 



Fig 5. 






Eanstetn'B Conetntctive Drawino. 



TANGENTIAL CIRCLES. 

78. — Fig. 1.— Problem — To divide the surface of a 
circle into three equivalent parts bo^mded by semi- 
circles. 
SohUion. — Let C be the given circle. Divide the 

diameter D A into 6 equal parts, and describe with 

1 and 5 as centers, 1 D as radius, the semicircles 

2 D and 4 A, with 2 and 4 as centers, and 2 D as 
radius, the semicircles D 4 and 3 A; D4A2 = Jof 
area of circle. 

79.— Fig. 2.— Problem.— lb construct a rosette of four 
xinits within a given circle. 

Solution. — Let A B be the diameter of the given 
circle. Draw four equal tangential circles within 
the given circle (See Figs. 5 and 6, Plate 18) and 
connect their centers by the lines F E, E D, D H, 
and H F, which at I, I', I", and I'" pass through 
their points of contact. 

Concentric arcs may be added to indicate material. 
80. — Fig. 3.— Problem. — To construct three tangential 
circles within a semicircle. 

Solution. — Let A D B be the given semi-circle. 
Divide the radius C D into 4 equal parts, erect at 
point 1, E F perpendicular to C D, and describe 
with C as center, and radius C 3, the arc E 3 F. 
Point 3 is center, 3 D the radius to circle C D, and 
E and F are the centers to the required tangential 
remaining circles. 

A and B are the centers, A B the radius to arcs 
A G and G B, which form a Gothic arch. 
81 — Pig. 4.— Problem. — To construct two semicircles 
andthrce circlestangential within a gice^i semicircle. 

Solution. — Let A 3 B be the given semicircle. 
Divide radius C 3 into 8, the diameter A B into 4 



equal parts; erect at E and F, E H and F G perpen- 
dicular to A B, and at 2, H G perpendicular to O 3. 
E F are the centers, E A the radius to semicircles 
A C and C B; 3 and 4 centers, 3—3 the radius to 
circles 3 — 4 and 4 — 3, and H and G the centers for 
the required remaining tangential circles. 

GOTHIC AND PERSIAN ARCHES. 

83.— Fig. 5.— Problem —To construct a Gothic arch 
on an equilateral triangle. {Inscribe a tangential 
circle. ) 

Solution. — Let A B be the equilateral triangle. 
Describe with B and A as centers, and radius A B 
the arcs A C and C B; A E H B is the required 
Gothic arch. 

Center F of a tangential circle in this arch is 
found by making D G = B A, D E = B G, and draw- 
ing E B, intersecting C G, in F; the center F and 
radius F G give the required tangential circle. 

Bemark. — When A R represents the thickness of 
the stone required in work, the arcs R S and S T 
are conceutric with A C and C B. The lines repre- 
senting the Joints of stones, as N B ( voussoir- 
lines ), are radii in the corresponding sector. 

83.— Pig. 6.— Problem.— To construct a Gothic arch 
when span and altitude are given. 
Solution. — Let A B be the given span and D E the 
altitude. Construct an isosceles triangle, A E B, 
with A B as base and D E as altitude; bisect A E 
by the perpendicular L I, which intersects span A B 
in I. I and K are the centers, I A the radius to arcs 
A E and E B. Make D F = A I, and P G = D I, 
and draw G I, intersecting D E, in H, the center, 
H D, the radius to the tangential circle in arch 
A EB. 



Fig. 1. 




Fig. 2. 




Plate 19. 

Fig. 3. 







R A f. 



U annUin i. CorulrucdDe Dravitng. 



84. — Fig. 1. — Solution 2.— Let A B be the span and 
C D the given altitude. Construct an isosceles 
triangle, A D B, in which the base = A B, the 
altitude = C D. Bisect A D by the perpendic- 
ular LI, intersecting the produced span in I; 
I and J are the centers, I A is the radius to 
arcs A D and B D. A D B is the required Gothic 
arch. To find center H for the inscribed circle, 
make CE = AI,EP = CI and draw F H I. 

85.— Fig. 2.— Problem.— To constnict a Gothic arch 
(wood or stone) with application of previous con- 
sti'uctions Jar its inside ornamentation. 
Heniark. — This problem is intended as a review 
of former constructions, and should be drawn not 
less than three times the size of Fig. 2, to avoid in- 
accurate work by crowded lines. 

86.— Pig. 3.— Problem.— To construct a Persian arch 

about an eciuilateral triangle. 
Solution. — Let A D B be the equilateral triangle. 
Divide A D into 3 equal parts and draw through 
point 2, parallel with D B, G 2 E, intersecting GH 
in G and A B in E. Make D H = D G, and draw 
H F parallel to D A. E and F are centers to arcs 
A 2 and B I, and G and H the centers to arcs 2 D 
and ID; A2DIBis the required Persian arch. 
87.— Fig. 4.— Problem.— To construct a Persian arch 

lehenAlB, the span, and CD, the altitude, are 

given. 



Solution. — Construct with span A B as base, and 
with altitude C D the isosceles triangle A D B. 
Trisect A D and erect in point 1 the perpendicular 
1 E; draw E 3 G, intersecting G H (parallel to A B) 
in G. Continue as in the previous construction and 
obtain the required Persian arch. 

egg-lines. 
88.— Fig. 5. — Problem. — To construct an egg-line on 
a given circle. 
Solution. — Let C be the given circle. Draw per- 
pendicular diameters A B and C D, also lines B D 
E and A D F ; B and A are centers, radius = A B 
to arcs A E and B F, and D center to arc E P ; 
A B P E is the required egg-line. To obtain a more 
elongated shape of an egg-line, place centers A' B' 
further out, but equidistant from C, and describe 
arcs A E' and B F', and with D as center arc E' F'. 

89.— Fig. 6. — Problem. — To constmct an egg-line 
when the short axis is given. 

Eemark. — The longest line possible to be drawn 
in the egg-line is called its long axis, and the 
greatest width perpendicular to it is the short axis. 

Solution. — Bisect the given short axis A B by 
the perpendicular D E, on which make H C J. 
CI I of A B; C P = C G = I of A B; F and G are 
centers, and P B the radius to arcs L N and K J, 
H to K L and ItoJN;KLNJis the required egg- 
line. 



Fig. 1. 




F,g. 2. 




Plate 20 

Fig. 3. 




Fig. 4. 



Fig 5. 



Fig 6. 








ffaimtcWB Consirnciivc Drawing. 



80.— Fig. 1.— Problem.— To construct an oval or lem-Une at ad- 
joining equal squares. 

Definition. — An oval is an elongated endless curve consisting 
of symmetric arcs. Tlie longest possible line drawn in an oval 
is called its Jong axis, and the greatest width perpendicular to 
it is called its short a.vis. Both axes divide the oval into sym 
metric parts. 

Solution.— Let A F Q C and P G D B be the given squares. 
Draw the diagonals F C and A Q, intersecting in H, and F D 
and BG, intersecting in I; G and F are centers, G A, the radius 
to arcs A B and C D, H and I the centers to arcs A C and B D. 

91.— Fio. 2.— Problem,— To consti'tict a»i oval at a given circle. 

Solution,— Let A B F Q be the given circle. Construct two 
perpendicular diameters, A F and B G, and draw A B D, A G I, 
F B E and F G H; F and A are the centers, radius F A to arcs 
E A H and D F I; B and G are the centers, radius B D to arcs 
E D and HI; E H I D is the required oval. 

92.— Fig. 3.— Problem,— To construct an oval, at two equal 

circles, of which the circumference of one passes through the 

center of the other. 

Solution.— Let A and be the given circles, intersecting each 

other in B and D. Draw from points B and D through centers 

A and C, lines BAG, BCH, DAF and D C B. D and B are 

the centers, radius D F to arcs F E, and G H. F B H G is the 

reauired oval. 

•3.- Fig. 4.— Problem.— To contitruct an oval when its long and 
short axes are given. 
Solution. — Let A B and D, bisecting perpendicularly, be the 
long and short axis respectively. Draw C B and the quadrant 
C K from center E. Make C N — K B and bisect N B by the I 



perpendicular O L H; I E = E H, and E J — L E, and draw 
H J P, IJ R and I L S. J and L are the centers, J A the radius 
to arcs P B and O S, and H and I the centers to arci F O and 
R S; P O S R is the required oval. 



94.— Figs. 6 and 6.— Problem.— To constr-uct an arch, its $pan 
and altitude beinff given. 
Solution.— het A B bo the given span, and C D, the perpen- 
dicular in its bisection point C, the altitude. Construct with H 
A B — A C and C D the rectangle C D E A, and draw diagonal 
D A. Bisect angles EDA and E A D by F D and F A. From 
F, perpendicular to A D, draw F H G and make I C = H C; 
H and I are the centers, with radius H A to arcs A F and 
B J, and 6 the center, radius G F to arcs FDJ; AFDJBlB 
the required arch. 

Remark.— When we assume the thickness of the stone used 
in the arch as B O, we describe the concentric arcs O N, N E 
and E L, and divide these into equal parts, except keystone K, 
to which generally more prominence is given. As in the Gothic 
arches, the joint lines of the stones are radii in the correspond- 
ing sector. 

95.— Fig. 6.— Solution 2.— Let A B be the given span, and C D the 
altitude. With E A as a radius shorter than the given 
altitude, and centers E, D and F, describe the circles E A. 
D G and F B; draw and bisect E G, by the perpendicular 
F H. The intersection of P H with the extended altitude, 
gives the center of arc I D L, whose radius is I H. Com- 
plete the required arch A I D L B and add its stone units. 



Fig. 1. 









/ s / 










V 




/ *> 


1 / \ 



Fig. 




Plate 21. 
Fig. 3. 




Fig. 5. 






Hanstcln's Conetnictivc Drolving. 



96.— Fig. I.— Solution 3.— Let A B be the span, and 
C D the altitude. Construct with A C the equi- 
lateral triangle A E C, and make C F = C D, 
and draw D F G. Parallel with E C dr aw G H I ; 
points H and K are the centers, A H the radius 
to arcs A G and J B, and I the center, I G the 
radius to arcs G D J. Proceed as in Sol. 2, and 
complete the required arch and its stone units. 

97.— Pig. 2.— Solution 4. — Let A B be the span, and 
O D the altitude. Construct on altitude C D 
the equilateral triangle DEC, make C F=C A, 
etc., complete similar to Fig. 1. 

98. — Pig. 4. — Problem.— To construct an elliptic arch 
when span and altitude are given. 

Solution. — Let A B be the given span, C D the 
altitude. Produce A B, and with radius C D' — 
C D^the given altitude describe semicircle J D'A. 
Divide J A and span A B similarly into the same 
number of equal parts, and erect at all division 
points perpendiculars. With the T square make CD 
=C' D', 2 E' and 4 E"=2 E, 1 G' and 5 G"=l G, etc.. 



and connect points B G" E" D E' G' A by a free- 
hand line, and complete the required elliptic arch. 

99. — Figs. 3 and 5. — Problem.— To construct ascend- 
ing arches when span and altitude are given. 

Solution 1. — Let A B be the given span and C B 
the altitude; draw C A, the ascending line. Make 
B D (the produced span)= B C, and bisect A D by 
the perpendicular EG; E is center, E A the radius 
to quadrant AG. F C is parallel to A B, F the 
center, F G the radius to quadrant G C. A G C B 
is the required arch. Complete and add the stone 
units as in previous constructions. 

100. — Fig. 5. — Solution^.. — Let AB be the given span, 
and B D the altitude. Draw D A, the ascending 
line, and bisect A B by the perpendicular F C: 
bisect angle F E A by G J, and with J as cen- 
ter, J A as radius, describe arc A P. Draw F J, 
then D H parallel to A B, and with center H, 
radius H D describe arc P D; A F D B is the 
required ascending arch. Complete and add 
stone units as in previous constructions. 



Flff. 1. 



'<'^' L 


J-J/>v 




"iC^ 



\ I// 



Fi^-- 2- 




Plate 22. 

Fig- 3. 



ff:>i 


— 



Fig: 4. 



.Fig. 5, 




AY,--^, U—-. 


o_ 


1 L-" 1 ~~~^ 





Hanslein'a CoiWfrucMoe Draifffnt^. 



IONIC SPIRALS. 

101. — Figs. 1 and 1 A.— Problenii— To construct an 
Ionic spiral when the altitude is given. 

Solution. — Let A B be the given altitude. Divide 
A B into 16 equal parts. The center of the spiral 
eye is situated in the 9th part from B, and its 
radius = ^ of A B. 

Pig. 1 A.—Bemarlc.—To explain division and sub- 
division, the eye of the spiral in double size is 
represented in Fig. 1 A. It is advisable to exe- 
cute Fig. 1, Plate 23 and Fig. 1, Plate 24 in as 
large a scale as possible, to facilitate an ac- 
curate division and subdivision. 

Draw vertical and horizontal diameters of the spiral e\c 
and upon these as diagonals the square D H G F. Draw 
then its diameters 1, C, 3 and 2. C. 4. Trisect the semi- 
diameters 1— C, 2 — C, etc., by the points 5 — 9.6 — 10, etc. 
Draw horizontal lines to the left, through points 1—2, 5—6, 
9—10; to the right through points 11—12. 7—8 and 3—4; then 
vertical lines downward through 2—3, 6—7, 10—11; upward 
through 12—9, 8-5 and 4—1. 

The spiral is composed of a series of Quadrants, whose 
vertices are the points, beginning with 12 down in order to 1, 
the last. The limits of each quadrant are determined by 
the vertical and the horizontal that start from the point in 
question as per above. Thus the verte.K of the first quadrant 
is point 12, its limits the lines 12— M and 12— I, its radius 
the distance of 12 from D. the beginning of the spiral. The 
second quadrant begins at I, its vertex is point ll, its limits 
the lines 11— I and 11— K, etc. There are three sets of four 
quadrants each. The second curve is similar to the first 
and can easily be understood if one remembers that for 
every point of the primary curve there is a corresponding 
point of the secondary curve. 

The set of centers of the secondary curve is found by 
trisecting the distances 12-C, 11— C, 10-C, 9— C, 8—12, 
9—11, etc., and using the point nearest the points 12, 11, 10, 
9, etc., in the same manner as in the primary curve. 



ELLIPTIC ASCENDING ARCHES. 

108.— Fig. 2.— Problem.— To construct an elliptic as- 
cending arch when span and altitude are given. 

Solution. — Let A B be the span and B C the alti- 
tude. Draw C A, the ascending line, and describe 
on A B as diameter, a semicircle, A NO B. Divide 
the diameter into any number of equal parts (6) and 
erect in each division perpendiculars, at which we 
make 2' N' = 2 N, 4' P = 4P and N' R' = N R and 
connect R' O' P' N', etc., by a free-hand line, which 
is the required arch. 

Bemarh. — This curve is also applied at the base 
of the Ionic column, as Fig. 6, Plate 9. 

103.— Fig. 3.— Problem.- Tb construct an elliptic as- 
cending arch when S2}an, its ascending and mean 
altitudes are given. 
Solution. — Let A B be the given span, B C the 
ascending and E F the mean altitude. With the 
mean altitude E P ^ E' F' describe the quadrant 
F' H A E'; divide radius E' A in 3 and subdivide 
the last 3d into 3 equal parts. Divide the span into 
the same number of proportional parts and erect 
perpendiculars. Transfer the altitudes of F' H J, 
etc., to the perpendicular A D, and draw lines 
parallel with the ascending line A C, to obtain the 
points of intersection J' J", H' H", F, etc., which 
points, connected by a free-hand line, will give the 
required arch, C J" H" F H' J' A. 




Fig. 1. 



Fig. 1. A 



Plate 23. 
Fig. 2. 



/ill 1^- 

I.H-T 1 




l' 1 1 ^ 

1 1 1 1 
1 III 






4 5 K 




HitTuKcin's CorwIiKiI.U'e Drniuino. 



.— FiGS.l and lA.— Solution.— This spiral differs 
from that of the preceding plate, in that its 
altitude A B is divided in 14 equal parts. The 
center of the spiral eye is the eighth point from 
B, and as in the previous case, its radius is 
equal to one of the divisions. The points 12, 
11, 10, 9, etc., are obtained by constructing the 
square D H E F (see Fig. lA), then draw its 
diameters 1 C 3 and 2 C 4. The semi-diameters 
are divided into three parts as follows: They 
are first bisected, and the part nearer C is again 
bisected. Now we are ready to start as in 
problem 101. 

The centers of the second curve, analogous 
to those of the primary are determined thus. 
Each innermost division of the semi-diameter 
is bisected for the centers of the first set of 
quadrants. Likewise the second division for 
the second set of four centers. For the remain- 
ing four divide the outer larger part of the semi- 
diameter into four equal parts and use the 
points nearest the ends of the diameter, i. e. 
1, 2, 3 and 4. 



105.— Fig. 3.— Problem.— To construct a spiral with 
semi-circles when the spiral "eye" is given. 

Solution. — Let C, a small circle, be the given 
spiral eye. Draw and produce a horizontal diam- 
eter, M A B N. With A as center, A B as radius, 
describe the semi-circle BO; C as center, C O as 
radius, semi-circle OP; A as center, A P as radius, 
semi-circle P R, etc. Curve B O P R, etc., is the 
required spiral. 



106.— Fig. 3.— Problem. 

a given triangle. 



-To construct the evolute of 



Solution. — Let A B C be the given triangle. Pro- 
duce C B, B A and AC; B is the center, B A the 
radius to arc A N; C the center, C N the radius to 
arc N O; A the center, radius A O to arc O P, etc., 
etc. Curve A N O P, etc., is the required e volute. 



Plate 24. 




HansUin's Constructive DrawinQ. 



CAM LINKS — AHCUIMEDEAN SPIRALS. 

Definition.'— An archimedean spiral is a curve in a plane 
generated by a point whjse distance from a centre of rotation 
increases uniformly. 

Cams are arrangements in mechanics by which a rotary 
motion is converted into a reciprocating action, they are con- 
structed by archimedean spirals. 

Remark. — The following curves, used principally in mechan- 
ics and architecture, should be executed by free-hand lines 
before the student attempts to use a curve rule. 

107. — Fig 1.— Problem.— To construct a cam-line of lU revolutions 
wlien the distance C C between revolutions Is given. 

Solution.— \fith 8 equal parts, 6 of which are equal to the 
given distance C C, describe the circle 8 A B D E F, which is 
divided into 6 equal parts by diameters. Describe circles with 
C as center, radius C J, to intersect diameter B F in B'; with 
radius C 2 to intersect D 8 in D'; C 3 to intersect E A in E', etc. ; 
connect points G B' D' E' F' 5 C H D by a free-hand line, to 
complete the required cam-line. 

108.— Fig. S.— Problem.— To construct a heart-shaped cam wlien the 
altitude is given. 

Bcmar/c.— Heart-shaped cams are made to convert half of a 
revolution into forward motion, the other half of the revolution 
into backward motion. ( Piston-rods for pumps, etc.) 

Solution —Let G 8 be the given altitude, which is divided into 
8 equal parts and is the radius, C the center of the circle, 
divided by diameters into 16 equal parts. With center C, radius 
C 1, describe circle to intersect radii C A and C G in A and A', 
with C 8 as radius to cut radii B C and J C in B' and B", with 
C 3 to cut radii G C and K C in C and O", etc. Connect 
CA' B' C D'l EHG" B" A" C by a free-hand line and complete 
the required heart-shaped cim. 

109. — Fig. 3.— Problem.— To construct a cam in 4 equal divisions, to 
raise a lever in the first a of its revolution, equal to the altitude 
B D, to remain stationary the second a, to descend its first 
position the thii-d %i. and remain stationary the last a of its 
revolution. 

SoluMon.- Let B D be the given altitude, B A an arbitrary 
tance from the hub, and C the center of the cam. Describe with 
C D, center C, the circle D H D' 4 and divide it into quadrants, 
two opposite ones Into 1 equal parts again, by diameters 
N 4 =• B D " the given altitudeis also divided iuto4equal parts, 
1, 2, 3 and 4, and with radius C 1 draw arcs 1 G', G, with radius 



C 3, 3 F' F, wit radius C 3, 3 E' E; connect N G' F' E' D' and 
the symmetric points B G F E H by a free-hand line and com- 
plete the required cam. 

110.— FiQ. 4.— Problem. To construct a cam in three equal divisions, 
which in one revolution shall lift a lever — A 4 in theffrst 3d, 
shall remain stationary the second 3d, and shall rise again the 
third 3d an altitude — 4 B and make a sudden escape at B, to 
renew its motion in the second re«oiu(ton . 

Solution.— het the two inner circles be shaft and hub circum- 
ferences. A 4 the altitude of the first incline, 4 B the altitude of 
the second incline (the third division). 

Remark. — This construction, in applying the principles of 
Figs. 1, 8 and 3, will not present any difficulty to the student, 



and ( 



1 now be solved without the assistance of a teacher. 



CONIC SECTIONS. ELLIPSE, PARABOLA AND HYPERBOLA. 

111. — Fig. 5, — Three curves, which we obtain by sectional planes 
through a circular cone and cylinder, are of the greatest im- 
portance in technical work; the ellipse, parabolaand hyperbola. 
A sectional plane through the cylinder or circular cone in an 
oblique direction, as U V or M N, respectively, creates the 
ellipse. A sectional plane S T, parallel to the side B of the 
circular cone, creates the parabola. A sectional plane Q R, 
parallel to the axis of the circular cone, creates the hyperbola. 
Definition.— An ellipse is a closed curve; the sum of the dis- 
tances of each point in this curve from two iixed points within, 
called foci, is equal to the long axis. The ellipse has two axes, 
the major and minor, bisecting each other perpendicularly and 
dividing the ellipse as well as its surface into two symmetric 
parts. 

112.— Fig. 6.— Problem.— I'o construct an ellipse when major axis 
(Iransversant) and minor axis (conjugant) are given. 

Solution.— Let A B be the major, C D the minor axis. With 
a radius H A B = A M and center C draw arc and intersections 
with A B, points F and F', the foci; divide F M arbitrarily into 
parts, increasing in length towards M, and with F and F' as 
centers, B 4 as radius, describe arcs E G and E' G'; with F and 
F' as centers, A 4 as radius, draw intersections at E and G and 
at E' and 6'. Points E E' G G' are situated at the circumference 
of the ellipse. Operate with points 3, 2 and 1 in the same 
manner, and we obtain by each operation 4 points, which lie at 
the circumference of the ellipse, as with point 2, e g,, by which 
we locate points H J H' J'. Connecting these points by a free 
hand line, we obtain C E H A J G, etc., the required ellipse. 



Fig. 1. 




Fig. 2. 










Plate 25. 
■Fig. 3. 




Fig. 4. 






Han&tcirCs Confiirwtfue I>rawinQ% 



H3.— FiQ.l — Problem.— To construct a tangent to an ellipie 
when the point of contact is given. 

Solution.— Let A B D be the ellipse and G the point of con- 
tact. Describe from G as center, withradius G F, the arc F N, 
and draw and produce line F' G, intersecting arc F N in N; bi- 
sect angle N Q F by IJ, which is the required tangent. 

Bcmark— In elliptic arches, executed in cut stone, the joints 
are perpendiculars (as P 6) to tangents, having the unit divisions 
as points of contact. 

lU.— Fig 1.— Problem.— F/'om an exterior point to construct a 
ta7igent to an ellipae. 
Solution.— liet H hi the given exterior point. With H as 
center, H F' as radius, describe arc F' O; with A B as radius, 
and F as center, intersect arc F' O in O. Bisect arc F' O by L H, 
which Is the required tangent. 

IIB.— Fig. 3, — Problem.— To constnict an ellipse when both axes 
are given. (Practical solution.) 

Solution 1. — Let A B and C D be the given axes. Find the foci 
(118) and place in F, F' and C pins, around which tie a linen 
thread to form the triangle F C F '. Take away the pin at C and 
place the pencil point in the triangle, by stretching the thread 
gently and forming a vertex of the triangle; draw the curve, 
which will be the required ellipse. 

Solution 2 —A B and C D are the given axes. Take O P, a 
straight edge or a slip of paper, at which make A' M' = A M = 
14A.B and A' C = C M = V4 C D. Guide the straight edge to 
have point C follow the major axis, and M' the minor axis, then 
will point A' describe the circumference of the required ellipse. 
Locate the position of point A' during this operation by pencil 
marks, which, connected, will give the ellipse. 

Bemo»/t.— Place In points C and M' pins, in point A' a pencil 
point, and let these pins slide in grooves In the place of the axes; 
we have an instrument called a trammel or ellipsograph, with 
which we are able to draw any ellipse by arranging points A' C 
and M' in the required proportions. 



116.— Fig. 3.— Problem.— To construct an ellipse by intersecting 
lines. 

Solution.- Let A B and C D be the given axis, and construct 
with these lines the rectangle E F G H; divide A B and E G into 
the same number of equal parts and number as in the diagram* 
Draw lines D 1 P, D 8 O and D 3 N, intersecting the lines C 1, CS 
and G 3 at P, O and N, etc., which points, connected by a free- 
hand line, will be the required ellipse. 

117.— Fig. *.— Problem.— To construct an elliptic curve in an 
oblique parallelogram. 

Solution.— Let B F G H be the parallelogram. Draw axes A B 
and C D bisecting opposite sides, and divide M and B C into 
the same number of equal parts; proceed as in the previous 
construction and draw C P O N A, etc., the required ellipse. 

118.— FiQ. 6 A.— Problem.— To construct an ellipse by intersec- 
tions of lines. 

Solution.— With A B and C D, the given axes, construct the 
rectangle B F H Q; divide E O and A E in the same number of 
equal parts (4) and number as shown In the diagram. Draw 
lines 1 A, 2 3, 3 2, and C 1, and connect their intersections T S R, 
etc., by a free-hand line to complete C T 8 R A, etc., the re- 
quired ellipse. 

119. — FiG.5 B.— Problem.— To construct an ellipse by its tangents. 

Solution.— Dt&w and divide C B into any number of equal 
parts (4): 1, 2, :i and 4, through which parallel with D draw 
P P', O O' and L L'; draw also E 1 1, E 2 K and ESN and lines 
LN,OKandPI, which are the tangents to the req uired ellipse. 
Draw the ellipse by a free-hand Une. 

120.— Fig. ().— Problem.— To construct an ellipse by the differ- 
ences of two circles. 

Solution, — Let B A and C D be the given axes. Describe with 
B A and C D as diameters concentric circles with center M. 
Divide both circles into 12 equal parts by the diameters 10, 4—11, 
B— 1, 7—2, 8 and 8, 9. Draw lines 7, 5—8, 4—10, 2 and 11, 1, and 
from the intersection points E F G and H the perpendiculars to 
10, 2—11, 1—7, 5 and 8, 4, which will give points N O A P K D, 
etc., at the circumference of the required ellipse. 



Fig. 1 




Fig. 2. 




Plate '26 
Fig 3. 





\ ^^^\ ""a5{i 







Fig. 4. 



Fig 5. 



Fig 5. 



Fig. 6. 






UanaUin'e Constrictive DrawinQ. 



PASABOLl. 

131.— Fig. 1.— Problem.— To construct a parabola when the axis 
and the base are gioen. 

Definition.— The parabola is a curve In which the distance of 
any point from an outside right line (direotrixj is equal to the 
distance of this point from a fixed point within, called focus. A 
line bisected perpendicularly by the asis at its terminus and 
Intersecting the curve is called the base, and a parallel with it, 
through the focus, the parameter of the parabola. 

Solution.- Let AP be the axis and LU the given base. Bisect 
LP = Vi the base L K in J, and draw J A. .In J erect a perpen- 
dicular to J A, J R intersecting the produced axis in R; trans- 
fer P R to left and right of point A, to obtain point F, the focus, 
and point O, through which draw M N, the directrix, perpen- 
dicular to the axis O P. Divide A P into arbitrary parts, 1, 2, 8, 
4, etc., in which erect perpendiculars, and with F as center, O 1 
as radius, cut the perpendicular 1 in B and B'; with O 2 as ra- 
dius, the same center, cut the perpendicular 3 in C and C ; with 
O 3 as radius cut perpendicular 3 in D and D', etc., and connect 
the obtained points L B' B' A B B K by a free-hand line, which 
is the required parabola. 

132.— Fig. 3,— Problem.— To construct a tangent to a parabola 
when the point of contact is given. 
Solution.- Let L RK be the given parabola, O P the axis, M N 
the directrix, and A the point of contact. With A as center, 
A F as radius, draw arc F B and A B perpendicular to M N. Bi- 
sect arc F B by line S G, which is the required tangent. 

Problem.— To construct a tangent to a parabola from an ex- 
terior point, R. 

Soiution.- With E as center, and B F as radius, draw arc F D 
and erect at D a perpendicular to M N, Intersecting the para- 
bola in H, the point of contact ; or bisect arc D F by line T E, 
which is the required tangent. 



123.— Fig. 3.— Problem.— I'o construct a parabola when two gym' 
metric tangents are given. 
Solution.— het B E = A E be the given tangents. Divide E B 
and A E into equal parts and number as shown in the diagram. 
Draw lines 7-7, 6-6, 6-5, 4-4, etc., which are the tangents of the 
parabola. A free-hand curve tangential to these tangents is the 
required parabola. 

124.— Fig. i.—Probltm.— To construct a parabola wTien the axis 
and the base are given or the rectangle drawn with these 
lines. 
Solution.— het A B 6 J be the given rectangle. Divide H 6 J 
= D 6 and B 6 into 6 equal parts, respectively ; number as in the 
diagram, and draw parallel to the axis D O lines through 1, 3, 3, 
4, 5. Draw also lines 5D, 4D,3D,3DandlD, intersecting with 
the horizontals in points I H G E F D, etc., which points, con- 
nected by a free-hand line, furnish the required parabola. 

125.— Fig. 5.— Problem.— To construct a parabola practically 
when base, O P, and axis, A B, are given. 
Solution.— Locate the focus F and the directrix M N and 
place a straight edge firmly coinciding with it. Fasten a thread 
to a pin placed in F and pass it around a pin in A to a point D 
of the set square, when its side G D coincides with axis A B. 
Remove the pin in A and hold the pencil to stretch the thread 
gently, touching C D constantly, shift the set square to the left. 
The pencil point will describe the required parabola on tha 
drawing paper. 

136.— Fig. 6.— Problem.— To construct a Oothic "rch by para- 
bolas. 
Solution.- Let A B be the span and F E the altitude of the 
arch. Construct the rectangle C D B A, divide C D into 8 and 
E F into i equal parts and number as the diagram. Draw lines 
1 A, 2 A, 3 A and parallel to span III', J 2 J', and H 3 H'. The 
points of intersection, A IJ H B H' J' I' B, connected by a free- 
hand line, complete the arch. 



Fig. 1. 




Fig. 2. 




Plate 27. 

Fig. 3. 




Fig. 4. 



Fig 5. 



Fig. 6. 



■^ 


^^^1 


^-^ 


^^:^_ 


rtrir-i 





BaneUin'a Comtruetiee DrawliUr- 



127. — Fig. 1.— Problem. — To construct hyperbolas 
when the vertices and foci are given. 

Definition. — The hyperbolas are curves; the dif- 
ference of distances of each point to the foci is 
equal to an invariable line, the axis. 

Solution. -Place on line M N, A and B the ver- 
tices, and F and F' the foci equidistant from O. 
From F' towards M mark arbitrary divisions and 
number as in diagram. With radius B 1, cent r 
F, — radius A 1 and center F' draw intersecting 
arcs at C and C; radius B 2, center F and radius 
A 3 and center P' draw intersecting arcs at D' and 
D, etc. Connect G' E' D' C A C D P G by a free- 
hand line, to complete the required hyperbola. To 
obtain the second curve, operate symmetrically. 
128. — Pig. 2.— Problem. — To construct a tangent to a 
hyperbola idhen point of contact, P, is given. 

Solution. — Draw line P F, and with radius P F' 
and center P the arc P' D. Bisect F' D by the line 
T U, which is the required tangent to the hy- 
perbola. 

Bemark. — The stone joints in hyperbolical arches 
are the perpendiculars to tangents at the point of 
contact. 

Problem. — From an exterior point, R, to construct a 
tangent to the hyperbola. 

Solution. — With R as center and radius R P 
draw arc P N; with P' as center and radius A B 



cut arc P N in N and bisect P N by S R, which is 
the required tangent to the hyperbola. 
139. — Pig. 3.— Problem — To construct hyperbolas 
when axis A B «s given; to find foci and draw the 
asymptotes. 

Asymptotes are right lines to which the branches 
of the hyperbolas approach when produced, 

but do not touch. 

Solution. — Construct the square EDCG with 
C D = A B, which the axis divides into two equal 
rectangles. Draw and produce the diagonals M N 
and O P, which are the required asymptotes. With 
O as center, O G as radius, draw arcs G P' and 
C F. With F and P', the required foci, draw the 
hyperbolas, as in Fig. 1. 

EVOLUTE. 

180.— Pig. 4.— Problem.— To construct an evolute at 
a given circle. 

Definition. — An evolute is a curve made by the 
end of a string unwinding from a cylinder. 

Solution. — Let C be the given circle ( the section 
of a cylinder). Divide the circumference into a 
number of equal parts (12) and draw the diameters 
and tangents 1 A', 3 A", 3 A'", 4 A*, etc. With 
center 1 and radius 1 A describe arc A A'; center 
2, radius 2 A', the arc A' A"; center 3, radius 
3 A", the arc A" A'", etc.; curve A, A', A", A'" is 
the required evolute. 



Fig. 1. 



Plate 28 

Fiff. 2. 




Fig. 3. 




Fig. 4. 




HavMeWt C&nstructiv€ Drawing. 



GEAR LINES— CYCLOID. 

131. — Fig. 1. — Problem. — To construct a cycloid 
when the generating point A is given at the cir- 
cumference of the circle. 

Definition.— k. cycloid is a curve generated by a 
point at the circumference of a circle, making one 
revolution in rolling on a straight line. The curve 
generated, when the circle rolls on the outside cir- 
cumference of another circle, is the epicycloid, and 
when the circle rolls on the inside circumference 
of another circle, the hypocycloid. 

Solution 1. — Let C be the rolling circle, tangent 
A B its rectified circumference and A the generat- 
ing point. Divide the circle C and line A B into 
the same number of equal parts (12) and number as 
in diagram. Pass horizontals through points 1, 2, . 
3, etc., of the rolling circle and erect perpendicu- 
lars at A B in points 1, 2, 3, etc. With points C, 
C", C^, C-" as centers, C A as a radius, describe 
circles 1 A', 2 A", 3 A'", 4 A*, etc., which points 
connected give the required cycloid. 

132. — Fig. 2.— Solution 2.— Follow the operations 
of the previous construction. Draw the circle 



C 6, also chords 6 1, 6 H, 6 G, G F, 6 E and 
their symmetric chords. Parallel to 6 E draw 
E' 7 K, to 6 F, F' 8 L', to 6 C, G' 9 M', to 6 H, 
H' 10 N' and to 6 I, I' 11 O'. 11 is the center, 
radius 11 B for arc B I', O' the center, radius 
O' I' to arc I' H', N' the center, radius N' H' 
for arc H' G', M' the center, radius M' G' to 
arc G' P', L' the center, radius L' F' to arc 
F' E', and K E' the radius to arc E' D E. Com- 
plete the construction symmetrically to the left 
of axis D K. The curve of B I' H' G', etc., is 
the required cycloid. 
When a cycloidal arch is executed in stone, the 

radii of the pertaining arcs are the joints of the 

units. 

133. — Fig. 3. — Problem.— To construct a cycloid 

when the point generating the curve is situated 

at a greater radius than that of the rolling circle. 

Solution. — Let C G be the rolling circle, G 12 its 

rectified circumference and A the generating point. 

Describe with C A from C a concentric circle and 

proceed in this construction as in Fig. 1. Pass 

horizontals through the divisions of the greater 

circle aud describe with radius C A and centers 

C, OS C\ etc., the circles B A', C A", D A'", etc. 

The curve passing through points A, A', A", A'", 

etc., is the required cycloid. 



Plate 29. 



Fig. 2. 




V" 



Fiff, 3. 




Han&iein's Conslructiot Drawing. 



GEAR LINES — EPICYCLOID AND HYPOCYCLOID. 

134. — Fig. 1. — Problem. — To construct an epicycloid 
when the relation of the rolling to the stationary 
circle is 1 : 2. 

Solution. — Let A B and 6 A be the diametei-s ol' 
tbe given circles, having the proportion of 2 : 1, 
respectively. Divide the rolling circle into any 
number of equal parts (12), and as circumferences 
are proportional to diameters, the circumference 
of 6 A = the semi-circumference B A contains 12 
of the same equal parts. With center C draw 
circles passing through points 1, 2, 3, 4, 5, 6 and 
D, and also the radii D'a, D"b, D"'c, etc. 
.D', D", D'" are the centers and radius D A to arcs 
aA', bA", cA'", etc. 

Connect A, A', A", A'", etc., by a curve, which 
is the required epicycloic. 



135. — FlG.2. — Problem. — To construct a hypocycloid- 

Solution. — Let C be the circle, point A the gener- 
ating point rolling in circle E. Relation of circles 
1 : 8. Make an equal division in both circles (A b 
= A 1) and draw radii A C, b C, c C", etc. C, C, 
C", C", etc., are the centers and C A the radius to 
arcs b A', c A", d A'", etc. Connect A, A', A", A'" 
by a curve, which is the required hypocycloid. 

136. — Fig. 3. — Problem, — To construct a. hypocycloid 
fvhen the relation of the circles is as 1 : 2. 

Solution. — Treating this construction as the pre- 
vious one, we shall obtain a right line A B as the 
required hypocycloid. 

This construction is the fundamental principle 
of the planet wheel, applied to convert directly a 
rotation into a reciprocating movement (pump- 
piston). 



137. — Pig. 1. — Problem. — To constntct an epitrochoid. 

It will not be difficult to execute this curve. See 
Fig. 2, Plate 29. 

138. — Fig. 2. — Problem. — Tv construct a hypotro- 
choid. 

Solution. — Let C F be the rolling circle, A the 
generating point and D J B the circumference on 
which circle C rolls. Proceeding as in Figs. 2 and 
3, we obtain the curve A, A', A", A'", etc., which 
is the required hypotrochoid. 



APPLICATIONS TO ARCHITECTURE. 

139.— Fig. 1.— Problem.— ro construct a design for 
an ornamented Gothic arch in stone. 

This construction is based on principles ex- 
plained and described in the previous part of this 
volume, and its solution should not present any 
serious difficulties to the student. 

Bemarlc. — To obtain an accurate result, it is ad- 
visable to make the equilateral triangle, the fun- 
damental figure of this arch, not less than 8 
inches a side. 



APPLICATIONS TO MECHANICS. 

140.— Fig. 1.— Preblem.— To construe* a pair of 
spur-wheels, their relation tohel : 2. 

To solve this problem we require the construc- 
tion of two epicycloids and two hypocycloids to 
the "flanks" of the teeth, and it is advisable to 
enlist the advice of a teacher, to execute this im- 
portant construction correctly- 



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